Answer
The solution is $x=2$
Work Step by Step
$\sqrt{x+\sqrt{x+2}}=2\hspace{0.7cm}{\color{blue}{\text{Given equation}}}$
$\Rightarrow (\sqrt{x+\sqrt{x+2}})^2=2^2$
$\Rightarrow x+\sqrt{x+2}=4$
$\Rightarrow \sqrt{x+2}=4-x$
$\Rightarrow (\sqrt{x+2})^2=(4-x)^2$
$\Rightarrow x+2=16-8x+x^2$
$\Rightarrow x^2-9x+14=0$
$\Rightarrow (x-7)(x-2)=0$
$\Rightarrow x-7=0$ or $x-2=0$
$\Rightarrow x=7$ or $x=2$
$\underline{\textbf{Check the answer}}$
At $x=7$
LHS $\sqrt{7+\sqrt{7+2}}=\sqrt{7+\sqrt{9}}=\sqrt{7+3}=\sqrt{10}\ne$RHS $\hspace{0.3cm}{\color{red}{Reject}}$
At $x=2$
$\sqrt{2+\sqrt{2+2}}=\sqrt{2+\sqrt{4}}=\sqrt{2+2}=\sqrt{4}=2=$RHS
The solution is $x=2$